If cot - 1 3 4 = x , find the value of cos x . (in the form of a b )

Given equation is, cot - 1 3 4 = x cot - 1 x = 3 4 = A P Using P.T. A C 2 = A B 2 + B C 2 A C 2 = 3 2 + 4 2 A C 2 = 9 + 16 A C 2 = 25 A C = 5 cos x = Base hypotenuse cos x = 3 5 . Hence, cos x value is 3 5 .

If tan - 1 3 4 = x , find the value of cos x and sin x . (in the form of a b )

Given equation is, tan - 1 3 4 = x tan x = 3 4 tan x = P B = 3 4 Using P.T. H 2 = A 2 + L 2 H 2 = 4 2 + 3 2 H 2 = 16 + 9 H 2 = 25 H = 5 sin x = L H sin x = 3 5 [ Substituting value from L and H ] cos x = Base hypotenuse cos x = 4 5 . Hence, the value sin x = 3 5 and cos x = 4 5 .

If cot - 1 - 1 7 = x , find the value of sin x and cos x .

Given equation is, cot - 1 - 1 7 = x cot x = - 1 7 cosec x = 1 + cot 2 x cosec x = 1 + - 1 7 2 cosec x = 1 + 1 49 cosec x = 49 + 1 49 cosec x = 50 49 cosec x = 5 2 7 cosec x = 1 sin x sin x = 7 5 2 cot x = cos x sin x cos x = cot x · sin x cos x = - 1 7 · 7 5 2 [ Substituting value from cot x and sin x ] cos x = - 1 5 2 sin x = 7 5 2 and cos x = - 1 5 2 .

If tan - 1 x = sin - 1 1 2 , find the value of x . (in the form of a b )

The given function is, tan - 1 x = sin - 1 1 2 ⇒ tan - 1 x = sin - 1 sin π 6 ∵ sin π 6 = 1 2 Using Properties of Inverse Trigonometric Function sin - 1 sin x = x ⇒ tan - 1 x = π 6 ⇒ x = tan π 6 ⇒ x = 1 3 . ∵ tan π 6 = 1 3 Hence, the value of 1 3 .

Evaluate the cot tan - 1 3 . (in the form of a b )

The given expression is, cot tan - 1 3 Using value from tan π 3 = 3 = cot tan - 1 tan π 3 Using formula tan - 1 tan x = x where x ∈ - π 2 , π 2 = cot π 3 Using value from cot π 3 = 1 3 cot π 3 = 1 3 . Thus, cot tan - 1 3 = 1 3 .

Prove the sin - 1 - 1 2 + cos - 1 - 3 2 = 2 π 3 .

We need to prove that, sin - 1 - 1 2 + cos - 1 - 3 2 = 2 π 3 L . H . S . = sin - 1 - 1 2 + cos - 1 - 3 2 = - sin - 1 1 2 + π - cos - 1 3 2 ∵ sin - 1 - x = - sin - 1 x ; cos - 1 - x = π - cos - 1 x = - sin - 1 sin π 6 + π - cos - 1 cos π 6 ∵ sin π 6 = 1 2 = - π 6 + π - π 6 ∵ sin - 1 sin x = x ; cos - 1 cos x = x = - 2 × π 6 + π = π - 2 π 6 = 6 π - 2 π 6 = 4 π 6 = 2 π 3 = R . H . S . Hence proved.